# Cauchy's Integral parametric conjugate

By considering the conjugate of its parametric form, evaluate $$\frac{1}{2\pi i}\int_{\gamma(0;1)}\frac{\overline{f(z)}}{z-a}dz$$ when $|a|<1$ and $|a|>1$, where $f$ is holomorphic in in the disk $(0;R), R>1$.

Typically when doing these kinds of integration and parametrization, $|z|=n$ is given, but it's different in this case (or is it not?). Can someone help me out?

• What is $\gamma(0;1)$? – kobe Mar 23 '15 at 2:31
• @kobe its the contour integral, centered at 0 and radius 1, if that answers your question? – rebc Mar 23 '15 at 2:34
• @science could you help me out how you obtained the answer? – rebc Mar 23 '15 at 2:40
• See here. – science Mar 23 '15 at 2:41
• The answer should be $\overline {f(0)}$. – science Mar 23 '15 at 2:43

Rewrite the integral as

$$\overline{\oint_{|z|=1} d\bar{z} \frac{f(z)}{\bar{z}-\bar{a}}}$$

Use the fact that $z \bar{z}=1 \implies d\bar{z} = -dz/z^2$; we get

$$-\frac1{a}\overline{\oint_{|z|=1} dz \frac{f(z)}{z \left (z-\frac1{\bar{a}}\right )}}$$

This is easily evaluated using the residue theorem. Note that the result depends on whether $a$ is within the unit circle. The result is

$$\frac1{i 2 \pi} \oint_{|z|=1} dz \frac{\overline{f(z)}}{z-a} = \begin{cases}\overline{f(0)} & |a| \lt 1 \\\overline{f(0)}- \overline{f \left (\frac1{\bar{a}} \right )} & |a| \gt 1 \end{cases}$$

• thanks for the clarification! – rebc Mar 23 '15 at 3:15